remove_if last character from a string - c++

Question

I would like to remove first and last brackets in passed by reference string. Unfortunately, I have difficulties with removing first and last elements conditionally.
I cannot understand why remove_if doesn't work as I expect with iterators.
Demo
#include <iostream>
#include <algorithm>
using namespace std;
void print_wo_brackets(string& str){
auto detect_bracket = [](char x){ return(')' == x || '(' == x);};
if (!str.empty())
{
str.erase(std::remove_if(str.begin(), str.begin() + 1, detect_bracket));
}
if (!str.empty())
{
str.erase(std::remove_if(str.end()-1, str.end(), detect_bracket));
}
}
int main()
{
string str = "abc)";
cout << str << endl;
print_wo_brackets(str);
cout << str << endl;
string str2 = "(abc";
cout << str2 << endl;
print_wo_brackets(str2);
cout << str2 << endl;
return 0;
}
Output
abc)
ac <- HERE I expect abc
(abc
abc

Answer 1

If remove_if returns end iterator then you will try to erase nonexistent element. You should use erase version for range in both places:
void print_wo_brackets(string& str){
auto detect_bracket = [](char x){ return(')' == x || '(' == x);};
if (!str.empty())
{
str.erase(std::remove_if(str.begin(), str.begin() + 1, detect_bracket), str.begin() + 1);
}
if (!str.empty())
{
str.erase(std::remove_if(str.end()-1, str.end(), detect_bracket), str.end());
}
}

Answer 2

The problem is here:
if (!str.empty())
{
str.erase(std::remove_if(str.begin(), str.begin() + 1, detect_bracket));
}
you erase unconditionally.
std::remove_if returns iterator to the beginning of range "for removal". If there are no elements for removal, it returns end of range (str.begin() + 1 in this case). So you remove begin+1 element, which is b.
To protect from this problem you shouldn't probably do something more like:
if (!str.empty())
{
auto it = std::remove_if(str.begin(), str.begin() + 1, detect_bracket);
if(it != str.begin() + 1)
str.erase(it);
}
I assume you simply want to check behavior of standard library and iterators, as otherwise check:
if(str[0] == '(' || str[0] == ')')
str.erase(0);
is much simpler.

Answer 3

Alternative:
#include <iostream>
#include <string>
std::string without_brackets(std::string str, char beg = '(', char end = ')') {
auto last = str.find_last_of(end);
auto first = str.find_first_of(beg);
if(last != std::string::npos) {
str.erase(str.begin()+last);
}
if(first != std::string::npos) {
str.erase(str.begin()+first);
}
return str;
}
using namespace std;
int main() {
cout << without_brackets("abc)") << endl
<< without_brackets("(abc") << endl
<< without_brackets("(abc)") << endl
<< without_brackets("abc") << endl;
return 0;
}
see: http://ideone.com/T2bZDe
result:
abc
abc
abc
abc

Answer 4

As stated in the comments by #PeteBecker, remove_if is not the right algorithm here. Since you only want to remove the first and last characters if they match, a much simpler approach is to test back() and front() against the two parentheses ( and ) (brackets would be [ and ])
void remove_surrounding(string& str, char left = '(', char right = ')')
{
if (!str.empty() && str.front() == left)
str.erase(str.begin());
if (!str.empty() && str.back() == right)
str.erase(str.end() - 1);
}
Live Example

Answer 5

All you need is this:
void print_wo_brackets(string& str){
str.erase(std::remove_if(str.begin(), str.end(),
[&](char &c) { return (c == ')' || c == '(') && (&c == str.data() || &c == (str.data() + str.size() - 1));}), str.end());
}
Live Demo
By stating:
str.erase(std::remove_if(str.end()-1, str.end(), detect_bracket));
You're evoking undefined behaviour.